5
$\begingroup$

Studying Taylor series, I wanted to get a sense for what higher derivatives really express in precise terms using the limit definition of the derivative.

Is this correct?

$$\frac{d^2y}{dx^2} = \lim_{h \to 0} \frac{\big(\lim_{k \to 0} \frac{f(x+h+k)-f(x+h)}{k}\big) - \big(\lim_{g \to 0} \frac{f(x+g)-f(x)}{g}\big)}{h}$$

$\endgroup$
  • 5
    $\begingroup$ yes it's correct, if you meant $\frac{d^2 f}{dx^2} =\frac{d f}{dx}(\frac{d f}{dx})$ and see wiki/Finite_difference#Higher-order_differences $\endgroup$ – reuns Jun 8 '16 at 20:24
  • $\begingroup$ @user1952009 Thanks for the link. I'm having trouble reconciling what is written there and what I have here, notably the $h^2$ they have in the denominator. $\endgroup$ – jeremy radcliff Jun 8 '16 at 20:26
  • 1
    $\begingroup$ it is because they let $g = k =h$ and you have to add a $\lim_{h \to 0} $ in front of the whole. now the limit doesn't depend on how $h \to 0$ (and if you let $g\to 0$ first, or $g = h$, or $g=h^2$) only when the function is twice differentiable (that's the definition : the limit exists, whatever how everything $\to 0$) $\endgroup$ – reuns Jun 8 '16 at 20:29
  • $\begingroup$ @user1952009 Ah! Thank you, this is very helpful. Also explains where the $(\Delta x)^2$ and higher degrees of delta $x$ come from in Taylor series. $\endgroup$ – jeremy radcliff Jun 8 '16 at 20:33
7
$\begingroup$

You are right, but you could do with an easier expression in one limit: $$ f''(x) = \lim_{h \to 0} \frac{f(x+2h) - 2f(x+h) + f(x)}{h^2} $$

$\endgroup$
  • $\begingroup$ Thank you, this makes sense now with @user1952009's comment. $\endgroup$ – jeremy radcliff Jun 8 '16 at 20:38
  • $\begingroup$ I fear that a coefficient $2$ is missing for the central term $f(x+h)$ as may be deduced from $\dfrac {\frac{f(x+2h)-f(x+h)}h-\frac{f(x+h)-f(x)}h}h$. The second order central difference (as linked by user1952009) $\dfrac {\frac{f(x+h)-f(x)}h-\frac{f(x)-f(x-h)}h}h=\dfrac{f(x+h)-2f(x)+f(x-h)}{h^2}\,$ may be better in practice (as well as $\dfrac{f\left(x+\frac h2\right)-f\left(x-\frac h2\right)}h$ for the derivative). $\endgroup$ – Raymond Manzoni Jun 8 '16 at 21:31
  • 1
    $\begingroup$ @RaymondManzoni indeed, a typo, corrected. thank you. You are right, centralized differences perform better, but here i chose this since it flows out of OP's formulation with $h=k$ $\endgroup$ – gt6989b Jun 8 '16 at 23:29
  • 3
    $\begingroup$ It should be pointed out that this is true only if the second derivative exists. The limit you give may exist even though the function is not twice differentiable at $x$. $\endgroup$ – symplectomorphic Jun 9 '16 at 5:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.